d Problem 6: The average weight of three man A, B and C is 84 kgs. A fourth man D joins them, and then average weight of the four becomes 80 kgs. If E , whose weight is 3 kgs more than that of D, replaces A. Then average weight of B, C, D and E becomes 79 kgs. Find weight of A. | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Problem 6: The average weight of three man A, B and C is 84 kgs. A fourth man D joins them, and then average weight of the four becomes 80 kgs. If E , whose weight is 3 kgs more than that of D, replaces A. Then average weight of B, C, D and E becomes 79 kgs. Find weight of A.

Problem 6: The average weight of three man A, B and C is 84 kgs. A fourth man D joins them, and then average weight of the four becomes 80 kgs. If E , whose weight is 3 kgs more than that of D, replaces A. Then average weight of B, C, D and E becomes 79 kgs. Find weight of A.

a)70 kgs             b) 80 kgs                           c) 75 kgs                         d) 72 kgs

Solution:

Let their be four persons and their weights be A, B , and D

As per statement average weight of A, B and C is 84 kg;

\Rightarrow \frac{A+B+C}{3}=84

\Rightarrow A+B+C=84\times 3

\Rightarrow A+B+C=252 ---(1)



After joining of 4th person their average weight becomes 80, i-e;

\Rightarrow \frac{A+B+C+D}{4}=80

\Rightarrow A+B+C+D=80\times 4

Put value of A+B+C from eq (1);

\Rightarrow 252+D=320

\Rightarrow D=320-252

\Rightarrow D=68

Now let another person E (whose weight is 3 kg more than D’s weight i-e E =D+3) replaces A– then by statement of problem

Average weight becomes 79;

\Rightarrow \frac{E+B+C+D}{4}=79

But ,

E=D+3=68+3=71

Put E=71 and D=68 in above;

\Rightarrow \frac{D+3+B+C+D}{4}=79

\Rightarrow \frac{2D+3+B+C}{4}=79

\Rightarrow \frac{2\times 68+3+B+C}{4}=79

\Rightarrow 139+B+C=79\times 4

\Rightarrow 139+B+C=316

\Rightarrow B+C=177

put value of B+C in eq (1);

\Rightarrow A+177=252

\Rightarrow A=252-177=75

So, A=75 Kgs

 

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